Electrical Power Engineering Intro 2

Massachusetts Institute of Technology

Department of Electrical Engineering and Computer Science

6.061 Introduction to Power Systems

Class Notes Chapter 2

AC Power Flow in Linear Networks ∗

J.L. Kirtley Jr.

1

Introduction

Electric power systems usually involve sinusoidally varying (or nearly so) voltages and currents. That is, voltage and current are functions of time that are nearly pure sine waves at fixed frequency. In North America, most ships at sea and eastern Japan that frequency is 60 Hz. In most of the rest of the world it is 50 Hz. Normal power system operation is at this fixed frequency, which is why we study how systems operate in this mode. We will deal with transients later. This note deals with alternating voltages and currents and with associated energy flows. The focus is on sinusoidal steady state conditions, in which virtually all quantities of interest may be represented by single, complex numbers. Accordingly, this section opens with a review of complex numbers and with representation of voltage and current as complex amplitudes with complex exponential time dependence. The dis­ cussion proceeds, through impedance, to describe a pictorial representation of complex amplitudes, called phasors. Power is then defined and, in sinusoidal steady state, reduced to complex form. Finally, flow of power through impedances and a conservation law are discussed. Secondarily, this section of the notes deals with transmission lines that have interesting behavior, both in the time and frequency domails.

2

Complex Exponential Notation

Start by recognizing a geometric interpretation for a complex number. If we plot the real part on the horizontal (x) axis√and the imaginary part on the vertical (y) axis, then the complex number z = x + jy (where j = −1) represents a vector as shown in Figure 1. Note that this vector may be represented not only by its real and imaginary components, but also by a magnitude and a phase ∗

c �2007 James L. Kirtley Jr.

1

y

Im(·)

z = x + jy �� � � � �

� �

� �

φ

�

�

�

�

x

Re(·)

Figure 1: Representation of the complex number z = x + jy

sin φ

Im(·)

� � �

�� � � � �

�

φ

Magnitude = 1

�

�

�

�

cos φ

Re(·)

Figure 2: Representation of ejφ angle: �

x2 + y 2 y φ = arctan( ) x

|z| =

(1) (2)

The basis for complex exponential notation is the celebrated Euler Relation: ejφ = cos(φ) + j sin(φ)

(3)

which has a representation as shown in Figure 2. Now, a comparison of Figures 1 and 2 makes it clear that, with definitions (1) and (2), z = x + jy = |z|ejφ

(4)

It is straightforward, using (3) to show that: ejφ + e−jφ 2 jφ e − e−jφ 2j

cos(φ) = sin(φ) =

(5) (6) (7)

2

Im(·)

z �� �

�

�

�

� �

�

� � � � � �

Re(·)

�

� �

�

�

�

� �� ∗ z

Figure 3: Representation Of A Complex Number And Its Conjugate The complex exponential is a tremendously useful type of function. Note that the product of two numbers expressed as exponenentials is the same as the exponential of the sums of the two exponents: ea eb = ea+b (8) Note that it is also true that the reciprocal of a number in exponential notation is just the exponential of the negative of the exponent: 1 (9) = e−a ea Then, if we have two numbers z 1 = |z 1 |ej φ1 and z 2 = |z 2 |ej φ2 , then the product of the two numbers is: (10) z 1 z 2 = |z 1 ||z 2 |ej(φ1 +φ2 ) and the ratio of the two numbers is:

|z | z1 = 1 ej(φ1 −φ2 ) z2 |z 2 |

(11)

z ∗ = x − jy

(12)

The complex conjugate of a number z = x + jy is given by:

The sum of a complex number and its conjugate is real: z + z ∗ = 2Re(z) = 2x

(13)

z − z ∗ = 2jIm(z) = 2jy

(14)

while the difference is imaginary:

where we have used the two symbols Re(·) and Im(·) to represent the operators which extract the real and imaginary parts of the complex number. The complex conjugate of a complex number z = |z|ejφ may also be written as: z ∗ = |z|e−jφ

(15)

so that the product of a complex number and its conjugate is real: zz ∗ = |z|ejφ |z|e−jφ = |z|2 3

(16)

3

Sinusoidal Time Functions

A sinusoidal function of time might be written in at least two ways: f (t) = A cos(ωt + φ)

(17)

f (t) = B cos(ωt) + C sin(ωt)

(18)

A third way of writing this time function is as the sum of two complex exponentials: f (t) = Xejωt + X ∗ e−jωt

(19)

Note that the form of equation 19, in which complex conjugates are added together, guarantees that the resulting function is real. Now, to relate equation 19 with the other forms of the sinusoidal function, equations 17 and 18, see that X may be expressed as: (20) X = |X|ejψ Then equation 19 becomes: f (t) = |X |ejψ ejωt + |X |∗ e−jψ e−jωt j(ψ+ωt)

= |X |e

∗ −j(ψ+ωt)

+ |X | e

= 2|X | cos(ωt + ψ)

(21) (22) (23)

Then, the coefficients in equation 17 are related to those of equation 19 by: A 2 ψ = φ

(25)

X = x + jy

(26)

|X| =

(24)

Alternatively, we could write in which the real and imaginary parts of X are: x = |X | cos(ψ)

(27)

y = |X | sin(ψ)

(28)

f (t) = x(ejωt + e−jωt ) + jy(ejωt − e−jωt )

(29)

Then the time function is written:

= 2x cos(ωt) − 2y sin(ωt)

(30)

A = 2x

(31)

B = −2y A B −j X = 2 2

(32)

Thus:

4

(33)

It is also possible to write equation 19 in the form: f (t) = Re(2Xejωt )

(34)

While both expressions (19 and 34) are equivalent, it is advantageous to use one or the other of them, according to circumstances. The first notation (equation 19) is the full representation of that sinusoidal signal and may be used under any circumstances. It is, however, cumbersome, so that the somewhat more compact version(equation 34) is usually used. Chiefly when nonlinear products such as power are involved, it is necessary to be somewhat cautions in its use, however, as we will see later on.

4

Impedance

Because it is so easy to differentiate a complex exponential time signal, such a way of representing time signals has real advantages in electric circuits with all kinds of linear elements. In Section 1 of these notes, we introduced the linear resistance element, in which voltage and current are linearly related. We must now consider two other elements, inductances and capacitances. The inductance

⊃ ⊃L ⊃ ⊃

C

Figure 4: Inductance and Capacitance Elements produces a relationship between voltage and current which is: vL = L

diL dt

(35)

If voltage and current are sinusoidal functions of time: v = V ejωt + V ∗ e−jωt i = Iejωt + I ∗ e−jωt Then the relationship between voltage and current is given simply by: V = jωLI

(36)

This is a particularly simple form, and as can be seen is directly analogous to resistance. We can generalize our view of resistance to complex impedance (or simply impedance), in which inductances have impedance which is: (37) Z L = jωL The capacitance element is similarly defined. A capacitance has a voltage-current relationship: i=C 5

dvC dt

(38)

Thus the impedance of a capacitance is: ZC =

1 jωC

(39)

The extension to resistive network behavior is now obvious. For problems in sinusoidal steady state, in which all excitations are sinusoidal, we may use all of the tricks of linear, resistive network analysis. However, we use complex impedance in place of resistance. The inverse of impedance is admittance: Y =

1 Z

Series and parallel combinations of admittances and impedances are, of course, just like those of conductances and resistances. For two elements in series or in parallel: Series: Z = Z1 + Z2 Y 1Y 2 Y = Y1+Y2

(40) (41)

Parallel:

4.1

Z1Z2 Z1 + Z2 = Y1+Y2

Z =

(42)

Y

(43)

Example

Suppose we are to find the voltage v(t) in the network of Figure 5, in which i(t) = I cos(ωt). The �� < � i(t) >R ��<

⊃ ⊃L ⊃ ⊃

+ v(t) −

Figure 5: Complex Impedance Network excitation may be written as: i(t) =

� � I jωt I −jωt e + e = Re Iejωt 2 2

Now, the complex impedance of the parallel combination of R and L is: R||jωL = So that, if v(t) is represented by: 6

RjωL R + jωL

V jωt V −jωt e + e 2 � 2� jωt = Re V e

v(t) =

Then

V =

RjωL I R + jωL

Now: the impedance Z may be represented by a magnitude and phase angle: Z = |Z |ejφ ωLR |Z| = � (ωL)2 + R2 π ωL φ = − arctan 2 R Then, using relations developed here, v(t) may be written as: ωLI v(t) = � � �2 cos(ωt + φ) 1 + ωL R Note that this expression represents only the sinusoidal steady state solution, and therefore does not represent any starting transients.

5

System Functions and Frequency Response

If we are interested in the behavior of a linear system such as the circuits we have been discussing, we often speak of the system function. This is the (usually complex) ratio between output and input of the system. System functions can express driving point behavior (impedance or its reciprocal, admittance) or transfer behavior. We speak of voltage or current transfer ratios and of transfer impedance (output voltage related to input current) and transfer admittance (output current related to input voltage). The system function may be expressed in a number of ways, often as a Laplace Transform. Such is beyond the scope of this subject. However, it is important to understand one way of expressing linear system behavior, in the form of frequency response. The frequency response of a system is the complex number that relates output of the system to input as a function of frequency. Usually it is expressed as a pair of numbers, magnitude and phase angle. Thus H(jω) = |H(jω)|ejφ(jω) Subjects in Signals and Systems or Network Theory often spend some time on how to obtain and plot the frequency response of a network in ways which are both useful and easy. For our purposes, a straightforward, perhaps even “brute force” approach will do. Consider, for example, the circuit shown in Figure 6. 7

1 mHy

� ∩∩∩∩

+ Vin − �

< 1

�

+ kΩ Vout − �

Figure 6: Example Circuit for Frequency Response This is just a voltage divider between an inductance and a resistance. We seek to find, and then plot, the transfer ratio Vout /Vin of this network. A very little analysis yields an expression for the transfer function, which is: 1 Vout (jω) R = = L Vin (jω) R + jωL 1 + jω R The magnitude and angle of this function can be extracted in a number of ways. For the pur­ pose of these notes, we have done the mathematics using MATLAB. The specific instructions for producing the frequency response plot are shown in Figure 7. Funamentally what is done is to compute the system function for a number of frequencies (note that we use a way of computing specific frequencies which produces a uniform spacing on a logarithmic scale, and then plotting the magnitude (also on a logarithmic scale) and angle of that system function against frequency.

6

Phasors

Phasors are not weapons. They are a handy geometric trick which help us understand the nature of sinusoidal steady state signals and systems. To start, consider the basis for complex exponential time notation, the function ejωt . At any instant of time, this is a complex number: at time t = 0 it is equal to 1, at time ωt = π2 it is equal to j, and so forth. We may describe this function as a vector, of length unity, rotating about the origin of the complex number plane, with angular velocity ω. It has, of course, both real and imaginary parts, which are just the projections of the vector onto the real and imaginary axes. Now consider a sinusoidally varying signal x(t), which may be represented by: x(t) =

X jωt X ∗ −jωt e + e 2 2

This is the sum of two numbers, complex conjugates, which are, as functions of time, rotating in opposite directions in the complex plane. The sum of the two is, of course, real. This is the same time function as: � � (44) x(t) = Re Xejωt

where the real part operator Re(·) simply takes the projection of the function on the real axis. It might be helpful at this point to remember one of the features of complex arithmetic. Mul­ tiplication of two complex numbers results in a third complex number which has: 8

L=1e-3; % Set Parameter Values R=1000; e=3:.05:7; % This is a way of producing evenly f=10 .^ e; % spaced points on a logarithmic chart om=2*pi .* f; % Frequency in radians per second H = 1 ./ (1 + j*L/R .* om); % This is the frequency response subplot(211); loglog(f, abs(H)) % Plot of magnitude xlabel(’Frequency, Hz’); ylabel(’Magnitude’); grid subplot(212); semilogx(f, angle(H)) % Plot of angle xlabel(’Frequency, Hz’) ylabel(’Angle’) grid title(’Frequency Response of L-R’)

print(’freq.ps’)

Figure 7: MATLAB Program freq.m 1. a magnitude which is the product of the magnitudes of the two numbers begin multiplied and, 2. an angle which is the sum of the angles of the two numbers being multiplied. Thus, multiplying a number by ejωt , which has a magnitude of unity and an angle which is increasing with time at the rate ω, simply has the effect of setting that number spinning around the origin of the complex plane. It is therefore relatively easy to represent sinusoidally varying signals with just their complex amplitudes, understanding that they also include ejωt , which provides time variation. The complex amplitude includes not only the magnitude of the signal, but also a phase angle. Usually the phase angle by itself is of little use, and must be related to some time reference. That is, as we will see, it is the difference between phase angles that is important in most cases. Impedances and admittances are also complex numbers, so that phasors can be used to visualize the relationship between voltages and currents in a network. The key here is that multiplication and division of complex numbers is the same as multiplication or division of magnitudes and addition or subtraction of angles.

6.1

Example

Consider the simple network shown in Figure 9, and suppose that the current source is sinusoidal: �

i = Re Iejωt

�

The impedance of the R-L combination is a complex number: 9

0

Magnitude

10

−1

10

−2

10

3

10

4

10

5

10 Frequency, Hz

Frequency Response of L−R

6

10

7

10

0

Angle

−0.5 −1 −1.5 −2 3 10

4

10

5

10 Frequency, Hz

6

10

7

10

Figure 8: Frequency Response Z = R + jωL = 1 + j2 Now: the impedance may be represented in the complex plane as shown in Figure 10. Voltage v is given by: � � v = Re V ejωt where:

V = ZI Then the relationship between voltage and current is as shown in Figure 11. Note that the phase angle between voltage and current is the same as the phase angle of the impedance. Note that KVL may be represented graphically in the fashion of Figure 12.

7

Energy and Power

For any terminal pair with voltage and current defined as shown in Figure 13, power flow into the element is: p = vi (45)

10

R = 1Ω ∧∧ + �� ∨ ∨ ⊃ ⊃ ωL = 2Ω � is v ⊃ ⊃ �� −

Figure 9: Example Circuit Im(Z) 2

� �

� �

� �

� �

Z

Re(Z) 1

Figure 10: Complex Impedance

Power is expressed in Watts (W), and one Watt is the product of one Volt and one Ampere. Energy transferred over an interval of time t0 to t1 is the integral of power: w=

� t1

v(t)i(t)dt

(46)

t0

Energy is expressed in Joules, and one Joule is one Watt- Second. A Joule is also a Newton-Meter (force times distance), and therefore a Watt is a Newton-Meter per Second. Consider the behavior of the three types of linear, passive elements we have encountered: • Resistance: v = Ri, Instantaneous power is: p = Ri2 = Im(·)

� �

� �

� �

� �

I

�

v2 R

V

Re(·)

Figure 11: Voltage and Current 11

(47)

Im(V )

�

� �

� �

V ��� � �

VL �

VR

Re(V )

Figure 12: Components of Voltage

i � + v −

element

Figure 13: Definition for Power di , Instantaneous power is: • Inductance: v = L dt

p = iL

di 1 di2 = L dt 2 dt

(48)

The quantity wL = 12 Li2 may be interpreted as energy stored in the inductance, so that L p = dw dt . We will need to refine this definition later, when we consider electromechanical interactions or nonlinear elements, but it will do for now. • Capacitance: i = C dv dt , Instantaneous power is: p = vC

dv 1 dv 2 = C dt 2 dt

(49)

The quantity wC = 12 Cv 2 may similarly be interpreted as energy stored in the capacitance. Next, consider the power input to each of these three elements under sinusoidal steady state conditions: • Resistance: if i = I cos(ωt + θ), then

p = RI 2 cos2 (ωt + θ) RI 2 = [1 + cos 2(ωt + θ)] 2 12

(50)

Thus, average power into the resistance is: 1 P = RI 2 2

(51)

• Inductance: if i = I cos(ωt + θ), then voltage is v = −ωLI sin(ωt + θ), and power is: p = −ωLI 2 cos(ωt + θ) sin(ωt + θ) ωLI 2 = − sin 2(ωt + θ) 2

(52)

Average power into the inductance is zero. Instantaneous energy stored in the inductance is 1 wL = LI 2 cos2 (ωt + θ) 2 and that has an average value:

1 < wL >= LI 2 4

(53)

• Capacitance: if v = V cos(ωt + φ), then i = −ωCV sin(ωt + φ), and power is: p=−

ωCV 2 sin 2(ωt + φ) 2

(54)

which has zero time average. Energy stored in the capacitance is:

1 wC = CV 2 cos2 (ωt + φ) 2 which has time average:

1 < wC >= CV 2 4

(55)

Now, consider power flow into a set of terminals in a situation in which both voltage and current are sinusoidal and have the same frequency, but possibly different phase angles: v(t) = V cos(ωt + φ) i(t) = I sin(ωt + θ) It is necessary to revert to the original form of complex notation, as in equation 19, to compute power. v(t) = i(t) =

� 1 � jωt V e + V ∗ e−jωt 2 � 1 � jωt Ie + I ∗ e−jωt 2

(56) (57)

Instantaneous power is the product of voltage and current: p=

� 1� ∗ V I + V ∗ I + V Iej2ωt + V ∗ I ∗ e−j2ωt 4

13

(58)

This is directly equivalent to:

� � 1 Re V I ∗ + V Iej2ωt 2

(59)

1 |V ||I| [cos(φ − θ) + cos(2ωt + φ + θ)] 2

(60)

p= This is, in turn, expressible as: p=

From this, we extract “real power”, or time- average power: 1 1 P = Re [V I ∗ ] = |V ||I| cos(φ − θ) 2 2

(61)

The ratio between real power and apparent power Pa = 21 |V ||I| is called the power factor, and

is simply: power factor = cos ψ = cos(φ − θ) (62)

The power factor angle ψ = φ − θ is the relative phase shift between voltage and current. This expression for time- average power suggests a definition for something we might call complex power: 1 P + jQ = V I ∗ (63) 2 in which average power P is the real part. The magnitude of this complex quantity is the apparent power. The imaginary part is called reactive power. It has importance which will be discussed later. Different units are used for real, reactive and apparent power, in order to gain some distinction between quantities. Usually we will express real power in watts (W) (or kW, MW,...). Apparent power is expressed in volt-amperes (VA), and reactive power is expressed in volt-amperes-reactive (VAR’s). To obtain some more feeling for reactive power, expand the time- varying part of the expression for instantaneous power: 1 pvarying = |V ||I| cos(2ωt + φ + θ) 2 Now, using the trig identity cos(x + y) = cos x cos y − sin x sin y, and assigning x = 2ωt + 2φ and y = −ψ = θ − φ, we have: 1 pvarying = |V ||I | [cos 2(ωt + φ) + sin ψ sin 2(ωt + φ)] 2 Thus, total instantaneous power is: p=

1 1 |V ||I | cos ψ [1 + cos 2(ωt + φ)] + |V ||I | sin ψ sin 2(ωt + φ) 2 2

(64)

Now, if we note expressions for P and Q, we can re-write this as: p = P [1 + cos 2(ωt + φ)] + Q sin 2(ωt + φ)

(65)

Thus, real power P represents not only time average power but also the pulsations that go with time average power. Reactive power Q represents energy exchange with zero average value.

14

7.1

RMS Amplitude

Note that, in all of the expressions for power used so far, a factor of 12 appears. This is, of course, because the average value of the product of two sinusoids of the same frequency has a value of half of the products of their peak amplitudes multiplied by the cosine of the relative phase angle. It has become common to use a different measure of voltage amplitude, which is called root-mean-square or simply RMS. The proper definition for the RMS value of a waveform is somewhat complex, but boils down to that value which, if it were DC, would dissipate the same power in a resistor. It is possible to define RMS for any periodic waveform. However, since we will be dealing with sinusoids, the definition is even easier. Clearly, since power dissipated in a resistor is, in terms of peak amplitudes: 1 |V |2 P = 2 R then the RMS amplitude must be: |V | VRM S = √ (66) 2 Then, V2 P = RM S R As we will see, RMS amplitudes are the default for most situations: when a circuit is described as “120 Volts A√ C”, the designation virtually always means 120 Volts, RMS. The peak amplitude of this is |V | = 2 · 120 ≈ 170 volts. Often you will see sinusoidal waveforms expressed in the form: √ v = 2VRM S cos(ωt) in which VRM S is obviously the RMS amplitude.

7.2

Example

Consider the simple network of Figure 14. We will calculate the instantaneous power flow into that network in terms we have been discussing. Assume that the voltage source has RMS amplitude ��

R ��< + −

v

⊃ ⊃X ⊃ ⊃

Figure 14: Example Circuit of 120 volts and R and X are both 100 Ω. Then: v(t) = 170 cos ωt The admittance of this network is: Y =

1 j − 100 100 15

Instantaneous Power Flow 350

300

250

Power, Watts

200

150

100

50

0

−50

−100

0

1

2

3

4 5 6 7 Frequency * Time, Radians

8

9

10

Figure 15: Power Flow For Example Circuit so that the complex amplitude of current is: I = 1.7 − j1.7 And then complex power is:

1 P + jQ = 170(1.7 + j1.7) 2 Real and reactive power are, respectively: P = 144 W, Q = 144 VAR. This gives a power factor angle of ψ = arctan(1) = 45◦ . Then, instantaneous power is: p = 144 [1 + cos 2(ωt − 45◦ )] + 144 sin 2(ωt − 45◦ )

This is illustrated in Figure 15.

8

A Conservation Law

It is possible to show that complex power is conserved in the same way as we expect time average power to be conserved. Consider a network with a collection of terminals and with a collection of internal branches. Instantaneous power flow into the network is: pin =

�

vi

terminals

Note that this expression holds for voltage and current expressed over any complete set of terminals. That is, if it is possible to delineate the terminals of the network into a set of pairs, the voltages might correspond to voltages across the pair, while currents would flow between the terminals of each pair. Alternatively, the voltages might correspond to single node-to-datum voltage, while 16

currents would then be single input node currents. Since power can go only into network elements, it follows that the sum of internal branch powers must be equal to the sum of terminal powers: �

�

vi =

terminals

vi

(67)

branches

If this is true for instantaneous power, it is also true for complex power: �

�

VI =

VI

(68)

branches

terminals

Now, if the network is made up of resistances, capacitances and inductances, �

�

VI =

resistances

terminals

�

VI +

�

VI +

inductances

VI

(69)

capacitances

For these individual elements: • Resistances:

V I ∗ = R|I|2

• Inductances:

V I ∗ = jωL|I|2 V I ∗ = −jωC|V |2

• Capacitances:

Then equation 69 becomes: �

�

VI =

R|I|2 + j

resistances

terminals

�

inductances

ωL|I|2 − j

�

capacitances

ωC |V |2

(70)

Then, identifying individual terms: �

VI =

2(P + jQ)

Total Complex Power into Network

terminals

�

resistances

j

�

R|I |2 =

2

ωL|I|2 = 4ω

inductances

j

�

capacitances

ωC|V |2 = 4ω

�

< pr >

�

< wC > Energy Stored in Capacitances

�

Power Dissipated in Resistors

< wL > Energy Stored in Inductances

So, for any RLC network: P + jQ =

�

resistors

9



< pr > +2jω 

�

inductors

< wL > −

�

capacitors



< wC >

(71)

Power Flow Through An Impedance

Consider the situation shown in Figure 16. This actually represents a number of important situ­ ations in power systems, where the impedance Z might represent a transmission line, transformer or motor winding. Of interest to us is the flow of power through the impedance. Current is given 17

il + vl � �� + −

V1

−

��

Z

+ −

��

V2

��

Figure 16: Power Flow Example by:

V1−V2 Z Then, complex power flow out of the left- hand voltage source is: il =

V ∗1 − V ∗2 1 P + jQ = V 1 2 Z∗ �

(72)

�

(73)

Now, the complex amplitudes may be expressed as: V 1 = |V 1 |ejθ

(74)

jθ+δ

V 2 = |V 2 |e

(75)

where δ is the relative phase angle between the two voltage sources. Complex power at the terminals of the voltage source V 1 is now given by: P + jQ =

|V 1 |2 |V 1 ||V 2 |e−jδ − 2Z ∗ 2Z ∗

(76)

This is describable as a circle in the complex plane, with its origin at |V 1 |2 2Z ∗ and its radius equal to: |V 1 ||V 2 | 2|Z| Now suppose the impedance through which we are passing power is describable as a simple inductance as shown in Figure 17. This is perhaps the simplest of transmission line models which represents only the inductive impedance of the line. Line inductance arises because currents in the line produce magnetic fields, and this is a fair model for most lines which are fairly ’short’. More on that in the next section. This line has the impedance Z = jωL = jXL Now, switching to RMS amplitudes, so that V s = reactive power flow are:

2V 1 and V r =

|Vs |2 − V s V ∗r Xl ∗ 2 V | | r − V sV r = −V r I ∗ = j Xl

Ps + jQs = V s I ∗ = j Pr + jQr

√

18

√

2V 2 , Then real and

I

L

Vs

Vr

Figure 17: Simplest Transmission Line Model Now if we assume that the voltages are of the form: V s = Vs ejφ V r = Vr ejθ and that the relative phase angle between them is φ − θ = δ and doing a little trig: Vs Vr sin δ XL Vs2 − Vs Vr cos δ = XL Vs Vr sin δ = − XL 2 Vr − Vs Vr cos δ = XL

Ps = Qs Pr Qr

A picture of this locus is referred to as a power circle diagram, because of its shape. Figure 18 shows the construction of a sending end power circle diagram for equal sending-end and receivingend voltages and a purely reactive impedance.

Q

V2 X

δ P

Figure 18: Power Circle, Equal Voltages As a check, consider the reactive power consumed by the line itself: we expect that Qs + Qr = QL , and so: Qs + Qr =

Vs2 + Vr2 − 2Vs Vr cos δ XL 19

Note that the voltage across the line element itself is found using the law of cosines (see Figure 19:

V

s

δ

V

l

V

r

Figure 19: Illustration of the Law of Cosines VL2 = Vs2 + Vr2 − 2Vs Vr cos δ

and, indeed,

QL =

10

VL2 XL

Compensated Line

L Cs

Vs

Cr

Vr

Figure 20: Transmission Line Model Perhaps a more commonly used model for a transmission line is as shown in Figure 20. This represents not only the fact that most transmission lines have, in addition to series inductance, parallel capacitance but also the fact that many transmission lines are shunt compensated. This may be represented as a two-port network with the admittance parameters, using XL = jωL and −j ,: XC = ωC Y ss

=

Y sr = Y rs = Y rr

=

1 1 − jXL jXC1 1 jXL 1 1 − jXL jXC2

It is fairly clear that, for voltage sources at both ends, real and reactive power flow are:

Ps =

Vs Vr sin δ XL 20

Qs Pr Qr

1 1 Vs Vr cos δ = − − XL XC1 XL Vs Vr sin δ = − X � � L Vs Vr cos δ 1 1 2 − − = Vs XL XC2 XL Vs2

�

�

Q

V

2

1

1

X L

X C

δ P

Figure 21: Power Circle, Equal Voltages, Compensation Offset The power circle for this sort of line is similar to that of the simpler model, but the center is offset to smaller reactive component, as shown in Figure 21. An interesting feature of transmission lines is illustrated by what might happen were the re­ ceiving line to be open: in that case: 1 Vr =Vs 1 − ω 2 LC Depending on the values of frequency, inductance and capacitance this could be arbitrarily large, and this is a potential problem, particularly for longer lines, as we will discuss in the next section.

11

Transmission Lines

A transmission line is really a long, continuous thing. It has inductance which is really inductance per unit length multiplied by the line length, but it also has a continuous capacitance. We might attempt to represent a long transmission line as a series of relatively ’short’ sections each represented by an inductance and a capacitance. These ’lumped parameter’ models for lines are actually used in many system studies, particularly in physical analog models called ’Transmission System Simulators’. (We built one of these at MIT in the 1970’s). After the next section you might contemplate the definition of ’short’ for our purposes here, but generally each lumped parameter capacitance and resistance pair would represent a few to a few tens of miles.

11.1

Telegrapher’s Equations

Peering at the model presented in Figure 22, one might divine that a proper representation of voltage and current, both of which are functions of time and distance along the line, might be: 21

L C

L C

L C

L C

Figure 22: Transmission Line Lumped Parameter Model

∂i ∂v = −Ll ∂x ∂t ∂i ∂v = −Cl ∂x ∂t These are known as the “Telegrapher’s Equations” and represent the fact that inductance presents voltage drop along the line in proportion to rate of change of current and that capacitance presents a change in current along the line in proportion to rate of change of voltage. It is not difficult to eliminate either voltage or current from these to produce a wave equation. For example, take the cross-derivatives and substitute the second of these equations into the first to get: ∂2v ∂2v = L C l l ∂x2 ∂t2 Now: this equation is solved by arbitrary functions which are of the form: v(x, t) = v(x ± ut) where the wave velocity is:

1 L l Cl So now we can see that the voltage on the line is the sum of some waveform going in the ’positive’ direction and something else going in the ’negative’ direction: u= √

v(x, t) = v+ (x − ut) + v− (x + ut) The same will be true of current, and substituting back into either of the telegrapher’s equations yields: i(x, t) =

1 (v+ (x − ut) − v− (x + ut)) Ll u

the product of inductance times wave velocity has the units of impedance: Ll u =

�

Ll = Z0 Cl

This is often referred to as the ’characteristic impedance’ of the transmission line. This is also a commonly used term: transmission cables are often referred to by their characteristic impedances. For coaxial wires 50 to 72 ohms are common values. For high tension transmission lines somewhat higher values are to be expected. 22

11.2

Surges on Transmission Lines

Consider the situation shown in Figure 23. Here the left-hand end of the line is driven by a current source with a pulse (illustrated is a square pulse). This is actually not too far from the situation that transmission lines experience with lightning, which is usually representable as a current source, typically of magnitude between 20 and 100 kA and duration of about 1µS. (Actually, it is not a square pulse but that is not important here). What will happen, if the pulse is short enough, is that it will launch a traveling wave in which v+ = Z0 i+ and i+ is the current that was imposed. When this pulse reaches the far, or load end of the line, we have the situation in which at that point: v(t) = v+ + v− v+ V− − i(t) = Z0 Z0 and, of coures, v = Ri. The ’reflected’, or negative going wave will have magnitude: R

v− = v+ ZR0 Z0

− 1

+ 1

In the extreme case of an open circuit, the magnitude of the voltage pulse at the end of the transmission line is exactly twice that of the propagating pulse. In the case of a short circuit, of course, the magnitude of the voltage is zero, the current in the short is double the current of the pulse itself, and the pulse is reflected, but going in the reverse direction with a polarity the opposite of the forward-going pulse. This is illustrated in cartoon form in Figure 23.

I

i

i

v

v

R > Z0

R

s

i

R < Z0 v

Figure 23: Pulse Propagation on a Transmission Line

11.3

Sinusoidal Steady State

Now, consider a transmission line operating in the sinusoidal steady state. As suggested by Fig­ ure 24, it is driven by a voltage source at one end and is loaded by a resistive load at the other. Consistent with the voltage and currents that we know can exist on such a line, we know they will be of this form: 23

V s

R

Figure 24: Transmission Line in Simple Configuration

�

v(x, t) = Re V + ej(ωt−kx) + V − ej(ωt+kx) i(x, t) = Re

�

�

V + j(ωt−kx) V − j(ωt+kx) e − e Z0 Z0

�

Where the phase velocity is u = ωk = √L1 C .

l l At the termination end of the line, at x = ℓ

R=

V e−jkℓ + V − ejkℓ V = Z0 + −jkℓ I V +e − V − ejkℓ

This may be solved for the ratio of ’reverse’ to ’forward’ amplitude: R

V − = V + e−2jkℓ ZR 0 Z0

−1

+1

Since at the ’sending’ end: Vs = V + + V − With a little manipulation it can be determined that e−jkℓ

V r = Vs �

R Z0

��

�

+1 +

�

+ 1 + e−2jkℓ

Further manipulation yields:

V r = Vs

R Z0

�

R Z0

�

R Z0

��

−1

�

−1

R Z0 R Z0

cos kℓ + j sin kℓ

This might be made a bit more comprehensible when turned into a magnitude: Vr | | = �� Vs R

Z0

R Z0 �2

cos kℓ

+ (sin kℓ)2

If the line is loaded with a resistance equivalent to the ’surge impedance’ (so-called ’surge impedance loading’, the receiving end voltage is the same as the sending end voltage. If it is more heavily loaded, the receiving end voltage is less than the sending end and if it is less heavily loaded the receiving end voltage is greater. 24

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